Question

triangle xyz was dilated by a scale factor of 2 to create triangle acb, yz = 5, and xz = 5.59. part a: find sin ∠x. explain how sin ∠x compares to sin ∠a. (5 points) part b: find cb and ba. you must show all work and calculations to receive full credit. (5 points)

Explanation:

Step1: Recall dilation properties

Dilation preserves angle - measures. So, $\angle X=\angle A$, and $\sin\angle X = \sin\angle A$. In right - triangle $XYZ$, if we assume $\angle Y = 90^{\circ}$ (since no other information about the right - angle is given and we need to find $\sin\angle X$), $\sin\angle X=\frac{YZ}{XZ}$.

Step2: Calculate $\sin\angle X$

Given $YZ = 5$ and $XZ=5.59$, then $\sin\angle X=\frac{YZ}{XZ}=\frac{5}{5.59}\approx0.894$.

Step3: Use dilation formula for side - lengths

Since the scale factor of dilation from $\triangle XYZ$ to $\triangle ACB$ is $k = 2$, if a side of $\triangle XYZ$ is $s_1$ and the corresponding side of $\triangle ACB$ is $s_2$, then $s_2=k\times s_1$.

Step4: Find $CB$

The side $CB$ corresponds to $YZ$. Given $YZ = 5$ and $k = 2$, then $CB=k\times YZ=2\times5 = 10$.

Step5: Find $BA$

The side $BA$ corresponds to $XZ$. Given $XZ = 5.59$ and $k = 2$, then $BA=k\times XZ=2\times5.59 = 11.18$.

Answer:

Part A: $\sin\angle X=\frac{5}{5.59}\approx0.894$. Since dilation preserves angle - measures, $\sin\angle X=\sin\angle A$.
Part B: $CB = 10$, $BA=11.18$